60 grams of sodium hydroxide is required to completely neutralize 49 grams of sulphuric acid, percentage purity of sodium hydroxide sample is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
66.6%
B
33.3%
C
54%
D
90%

Solution:

$2 NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + 2 H_{2} O; (2 \times 40) g 98 g 80 grams of pure NaOH is required to neutralize 98 grams of H_{2} {SO}_{4}; 40 grams of pure NaOH is required to neutralize 49 grams of H_{2} {SO}_{4}; But in the given question, 60 grams of NaOH was required; Percentage purity of NaOH = \frac{40}{60} \times 100 = 66.6 %$

Solution:

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